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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Given a function <span class="process-math">\(u(x,t)\)</span> defined for all <span class="process-math">\(t &gt; 0\)</span> and assumed to be bounded we can apply the Laplace transform in <span class="process-math">\(t\)</span> considering <span class="process-math">\(x\)</span> as a parameter.</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}[u(x, t)]=\int_0^{\infty} e^{-st} u(x, t) \mathrm{d} t=U(x, s).
\end{equation*}
</div>
<p class="continuation">In applications to PDEs we need the following:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
{\mathcal L}[u_t(x, t)]&amp;=\int_0^{\infty} e^{-st} u_t(x, t) \mathrm{d} t=e^{-st} u(x, t)\big|_0^{\infty}+s \int_0^{\infty} e^{-st} u(x, t) \mathrm{d} t\\
&amp;=s U(x, s)-u(x, 0),
\end{aligned}
\end{equation*}
</div>
<p class="continuation">so we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}[u_t(x, t)]=s U(x, s)-u(x, 0).
\end{equation*}
</div>
<p class="continuation">In exactly the same way we obtain</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}[u_{tt}(x, t)]=s^2 U(x, s)-s u(x, 0)-u_t(x, 0).
\end{equation*}
</div>
<p class="continuation">We also need the corresponding transforms of the <span class="process-math">\(x\)</span> derivatives:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}[u_x(x, t)]=\int_0^{\infty} e^{-st} u_x(x, t) \mathrm{d} t=U_x(x, s),
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\mathcal L}[u_{xx}(x, t)]=\int_0^{\infty} e^{-st} u_{xx}(x, t) \mathrm{d} t=U_{xx}(x, s).
\end{equation*}
</div>
<span class="incontext"><a href="sec8_6.html#p-502" class="internal">in-context</a></span>
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